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Sphinxa [80]
2 years ago
14

What are some of the skills required to play floor hockey?​

Physics
2 answers:
yawa3891 [41]2 years ago
5 0
Passing, receiving passes, shooting, goal tending, stick handling and defensive skills
taurus [48]2 years ago
4 0

Answer:

Skills required to play floor hockey include passing, receiving passes, shooting, stick- handling (dribbling), defensive skills and goaltending. Players are allowed to use both sides of the blade of the stick. The front side is called the forehand (face) and the back side is called the backhand.

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In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
3 years ago
Why is physics used to study stars?
goblinko [34]

Answer:

stars are made of matter

3 0
2 years ago
Help !!!!!!!!<br>Please its Urgent.<br>​
Wewaii [24]

None can.

A clinical thermometer only measures temperatures above +30°C.

Mercury and alcohol are both frozen solid at -50°C.

6 0
2 years ago
At what angle should the axes of two polaroids be placed so as to reduce the intensity of the incident unpolarized light to 13.
prohojiy [21]

Answer:

θ = 66.90°

Explanation:

we know that

I= \frac{I_0}{2}cos^2\theta

I= intensity of polarized light =1

I_o= intensity of unpolarized light = 13

putting vales we get

1= \frac{13}{2}cos^2\theta

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therefore θ = 66.90°

5 0
3 years ago
A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo
statuscvo [17]
The Young modulus E is given by:
E= \frac{F L_0}{A \Delta L}
where 
F is the force applied
A is the cross-sectional area perpendicular to the force applied
L_0 is the initial length of the object
\Delta L is the increase (or decrease) in length of the object.

In our problem, L_0 = 3.0 m is the initial length of the column, E=3.0 \cdot 10^{10}N/m^2 is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m
and the cross-sectional area is
A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2
The force applied to the column is the weight of the load:
W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N

Now we have everything to calculate the compression of the column:
\Delta L =  \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m
So, the column compresses by 1.83 millimeters.
3 0
3 years ago
Read 2 more answers
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